Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{x + 5}{x + 10} \times \dfrac{-x^3 - 8x^2 + 20x}{x^3 + 7x^2 + 10x} $
First factor out any common factors. $z = \dfrac{x + 5}{x + 10} \times \dfrac{-x(x^2 + 8x - 20)}{x(x^2 + 7x + 10)} $ Then factor the quadratic expressions. $z = \dfrac {x + 5} {x + 10} \times \dfrac {-x(x + 10)(x - 2)} {x(x + 5)(x + 2)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {(x + 5) \times -x(x + 10)(x - 2) } {(x + 10) \times x(x + 5)(x + 2) } $ $z = \dfrac {-x(x + 10)(x - 2)(x + 5)} {x(x + 5)(x + 2)(x + 10)} $ Notice that $(x + 5)$ and $(x + 10)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-x(x + 10)(x - 2)\cancel{(x + 5)}} {x\cancel{(x + 5)}(x + 2)(x + 10)} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $z = \dfrac {-x\cancel{(x + 10)}(x - 2)\cancel{(x + 5)}} {x\cancel{(x + 5)}(x + 2)\cancel{(x + 10)}} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $z = \dfrac {-x(x - 2)} {x(x + 2)} $ $ z = \dfrac{-(x - 2)}{x + 2}; x \neq -5; x \neq -10 $